What is the equation of the line tangent to # f(x)=2xcos3x# at # x=pi/3#?

1 Answer

Tangent line:
#y=-2x#

Explanation:

From the given trigonometric equation

#f(x)=2x*cos 3x#

Let #y=2x*cos 3x#
solve for the point #(x_1, y_1)# of which #x_1=pi/3#

#y_1=2(pi/3)*cos 3(pi/3)=(2pi)/3*cos pi=(2pi)/3*(-1)#

#y_1=-(2pi)/3#

The point is #(pi/3, -(2pi)/3)#

Solve for the slope #m=y'(pi/3)#

#y'=2*[x*d/dx(cos 3x)+(cos 3x)d/dx(x)]#
#y'=2*[x*(-3*sin 3x)+(cos 3x)(1)]#

slope #m=y'(pi/3)=2*[(pi/3)*(-3*sin 3(pi/3))+(cos 3(pi/3))(1)]#

#m=2[-pi*sin pi+cos pi]#
#m=2[0-1]#
#m=-2#

Use Point-Slope form to find the equation of the tangent line

#y-y_1=m(x-x_1)#

#y--(2pi)/3=-2(x-pi/3)#

#y+(2pi)/3=-2x+(2pi)/3#

#y=-2x#

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God bless....I hope the explanation is useful.